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北仑网站建设网站,外贸企业招聘,创新的网站,帮人家做网站维护实用数学手册(v2)-1.1.8:合分比定理证明设 kaba−bk \frac{a b}{a - b}ka−bab​#xff0c;则 abk(a−b)a b k(a - b)abk(a−b)。 化简得到#xff1a; abka−kba b ka - kbabka−kb#xff0c; a−ka−b−kba - ka -b - kba−ka−b−kb#xff0c; a(1−k)−b(1k)…实用数学手册(v2)-1.1.8:合分比定理证明设kaba−bk \frac{a b}{a - b}ka−bab​则abk(a−b)a b k(a - b)abk(a−b)。化简得到abka−kba b ka - kbabka−kba−ka−b−kba - ka -b - kba−ka−b−kba(1−k)−b(1k)a(1 - k) -b(1 k)a(1−k)−b(1k)ab1k1−k\frac{a}{b} \frac{1 k}{1 - k}ba​1−k1k​。同样地设mcdc−dm \frac{c d}{c - d}mc−dcd​则cdm(c−d)c d m(c - d)cdm(c−d)。化简得到cdmc−mdc d mc - mdcdmc−mdc−mc−d−mdc - mc -d - mdc−mc−d−mdc(1−m)−d(1m)c(1 - m) -d(1 m)c(1−m)−d(1m)cd1m1−m\frac{c}{d} \frac{1 m}{1 - m}dc​1−m1m​。根据题设aba−bcdc−d\frac{a b}{a - b} \frac{c d}{c - d}a−bab​c−dcd​即kmk mkm。因此1k1−k1m1−m\frac{1 k}{1 - k} \frac{1 m}{1 - m}1−k1k​1−m1m​。由ab1k1−k\frac{a}{b} \frac{1 k}{1 - k}ba​1−k1k​和cd1m1−m\frac{c}{d} \frac{1 m}{1 - m}dc​1−m1m​且kmk mkm可以得出abcd\frac{a}{b} \frac{c}{d}ba​dc​。交叉相乘得到adbcad bcadbc再同时加上acacac且同时减去acacac即adacbcacadac bcacadacbcacacbd−acadacbd−acbcacbd-acad acbd-acbcacbd−acadacbd−acbc整理得到abbcdd\frac{ab}{b}\frac{cd}{d}bab​dcd​同理交叉相乘得到adbcad bcadbc再同时加上bdbdbd且同时减去bdbdbd即adbdbcbdadbd bcbdadbdbcbdadbd−bdabbcbd−bdabadbd-bdab bcbd-bdabadbd−bdabbcbd−bdab整理得到a−bbc−dd\frac{a-b}{b}\frac{c-d}{d}ba−b​dc−d​两式相比即可得到aba−bcdc−d\frac{a b}{a - b} \frac{c d}{c - d}a−bab​c−dcd​。所以合分比定理得汪。